Question

What is the derivative of (ln(x))^x ?

I have:

f(x) = ln(x)^x
f(x) = xlnx
f'(x) = x/x + 1 * ln(x)
f'(x) = 1 + ln(x)

Is this correct?

Answer 1

not quite, your brackets are very important
what you have worked out is ...
ln (x^x)
which is x lnx etc, and your answer would be correct for that

but you typed (lnx)^x
which would be (lnx)(lnx)(lnx) ... (lnx) , x times

we might apply the rule: extension of the product rule
if y = abcd..
y' = abc(d').. + abd(c').. + bcd(a').. + ....
that is, pick out one of the factors, differentiate that one, times the rest of them,
repeat for all factors.
so if you have n factors, there would be n terms in your derivative
so if
y = lnx lnx lnx ... lnx, x of them
y' = (1/x)(lnx)(x-1) + (1/x)(lnx)^(x-1) + .. same result x times
so y' = x(1/x)(lnx)^(x-1)
= (lnx)^(x-1)

I don't like this result at all

Answer 2

f = (ln x)^x
ln f = x ln(lnx)
1/f f' = ln(lnx) + x(1/lnx)*1/x
f' = (lnx)^x * (ln(lnx) + 1/lnx)

This agrees with the extension of the normal power/exponent rule. If

y = u^v where u and v are functions of x, then

y' = v u^(v-1) u' + lnu * u^v v'
= u^v (v/u u' + lnu v')

Here u(x) = lnx and v(x) = x
So, u' = 1/x and v' = 1

y' = (lnx)^x (1/lnx + ln(lnx))

Note that if u or v is a constant, we have
d/dx(u^n) = n u^n-1 u'
d/dx(a^v) = lna a^v v'

Answer 3

This is great, thank you!